Abstract algebra , intersection of ideals

[okoolo]

Homework Statement

prove that <x^m> intersection <x^n> = <x^LCM(m,n)>

Homework Equations

The Attempt at a Solution

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let b be in <x^n> intersection <x^m>

then for some t,k,p in Z, b=x^(mt) = x^(nk) thus b=x^(LCM(m,n) * p i.e. b is in <x^LCM(m,n)>

<===

let b be in <x^LCM(m,n)>
let s=lcm(m,n) =mt=nk for some t,k in Z

then for some r in Z , b=x^(sr)

then b = x^(mt)r =x^(nk)r

and that's where I'm stuck..

I was also considering trying to find and isomorphic mapping from one set to another..

any ideas?
thanks, Adam

 

[mighty2000]

Dear Adam,

Thank you for your post. To prove that <x^m> intersection <x^n> = <x^LCM(m,n)>, we can use the definition of intersection and the properties of exponents.

First, let's define <x^m> and <x^n> as follows:

<x^m> = {x^am | a is an integer}
<x^n> = {x^bn | b is an integer}

Then, the intersection of these two sets can be written as:

<x^m> intersection <x^n> = {x^c | c is an integer and x^c is in both <x^m> and <x^n>}

Now, let's consider an element x^c in the intersection. This means that x^c is in both <x^m> and <x^n>. So, there exist integers a and b such that x^c = x^am and x^c = x^bn.

Using the properties of exponents, we can rewrite x^c as x^(am+bn). Now, since c is an integer, we can let c = LCM(m,n) * p for some integer p. This means that:

x^c = x^(LCM(m,n) * p) = x^(am+bn)

Since am = bn, we can rewrite this as:

x^c = x^(LCM(m,n) * p) = x^(am+am) = x^(2am)

Since a and p are both integers, 2am is also an integer. Therefore, x^c is in <x^LCM(m,n)>.

This shows that any element in the intersection is also in <x^LCM(m,n)>. Now, we need to show that any element in <x^LCM(m,n)> is also in the intersection. This can be done using a similar argument as above, but in the opposite direction.

Therefore, we have shown that <x^m> intersection <x^n> = <x^LCM(m,n)>.

I hope this helps. Please let me know if you have any further questions. Keep up the good work in your studies!